Induction: One Step At A Time

Induction: One Step At A Time

Today we will talk about INDUCTION

Induction is the primary way we: • Prove theorems • Construct and define objects

Let’s start with dominoes

Domino Principle: Line up any number of dominos in a row; knock the first one over and they will all fall.

n dominoes numbered 1 to n • Fk ´ The kth domino falls • If we set them all up in a row then we know that each one is set up to knock over the next one: • For all 1 ≤ k < n: • Fk) Fk+1

n dominoes numbered 1 to n • Fk ´ The kth domino falls • For all 1 ≤ k < n: • Fk) Fk+1 F1) F2) F3) … F1) All Dominoes Fall

n dominoes numbered 0 to n-1 • Fk ´ The kth domino falls • For all 0 ≤ k < n-1: • Fk) Fk+1 F0) F1) F2) … F0) All Dominoes Fall

The Natural Numbers •  = { 0, 1, 2, 3, . . .}

The Natural Numbers •  = { 0, 1, 2, 3, . . .} One domino for each natural number: 0 1 2 3 4 5 ….

Standard Notation/Abbreviation“for all” is written “8” • Example: • For all k>0, P(k) • is equivalent to • 8k>0, P(k)

n dominoes numbered 0 to n-1 • Fk ´ The kth domino falls • 8 k, 0 ≤ k < n-1: • Fk) Fk+1 F0) F1) F2) … F0) All Dominoes Fall

Plato: The Domino Principle works for an infinite row of dominoes Aristotle: Never seen an infinite number of anything, much less dominoes.

Plato’s DominoesOne for each natural number • An infinite row, 0, 1, 2, … of dominoes, one domino for each natural number. • Knock the first domino over and they all will fall. • Proof: • Suppose they don’t all fall. Let k > 0 be the lowest numbered domino that remains standing. Domino k-1 ≥ 0 did fall, but k-1 will knock over domino k. Thus, domino k must fall and remain standing. Contradiction.

The Infinite Domino Principle Fk ´ The kth domino will fallAssume we know that for every natural number k, Fk) Fk+1 F0) F1) F2) … F0) All Dominoes Fall

Infinite sequence of statements: S0, S1, … Fk´ “Sk proved” Infinite sequence ofdominoes. Fk´ “domino k fell” Mathematical Induction: statements proved instead of dominoes fallen Establish 1) F0 2) For all k, Fk) Fk+1 Conclude that Fk is true for all k

Inductive Proof / ReasoningTo Prove k 2, Sk Establish “Base Case”: S0 Establish that k, Sk) Sk+1 • Assume hypothetically that • Sk for any particular k; • Conclude that Sk+1 k, Sk) Sk+1

Inductive Proof / ReasoningTo Prove k 2, Sk Establish “Base Case”: S0 Establish that k, Sk) Sk+1 • “Induction Hypothesis” Sk • “Induction Step” Use I.H. to show Sk+1 k, Sk) Sk+1

Inductive Proof / ReasoningTo Prove k ≥ b, Sk Establish “Base Case”: Sb Establish that k ≥ b, Sk) Sk+1 Assume k ≥ b “Inductive Hypothesis”: Assume Sk “Inductive Step:” Prove that Sk+1 follows

Theorem? The sum of the first n odd numbers is n2.

Theorem:? The sum of the first n odd numbers is n2. Check on small values: 1 = 1 1+3 = 4 1+3+5 = 9 1+3+5+7 = 16

Theorem:? The sum of the first n odd numbers is n2. The kth odd number is expressed by the formula (2k – 1), when k>0.

Sn “The sum of the first n odd numbers is n2.” Equivalently, Snis the statement that: “1 + 3 + 5 + (2k-1) + . . +(2n-1) = n2 ”

Sn “The sum of the first n odd numbers is n2.” “1 + 3 + 5 + (2k-1) + . . +(2n-1) = n2”Trying to establish that: 8n ≥ 1 Sn • Assume “Induction Hypothesis”: Sk • (for any particular k¸ 1) • 1+3+5+…+ (2k-1) = k2 • Add (2k+1) to both sides. • 1+3+5+…+ (2k-1)+(2k+1) = k2 +(2k+1) • Sum of first k+1 odd numbers = (k+1)2 • CONCLUDE: Sk+1

Sn “The sum of the first n odd numbers is n2.” “1 + 3 + 5 + (2k-1) + . . +(2n-1) = n2”Trying to establish that: 8n ≥ 1 Sn • In summary: • 1) Establish base case: S1 • 2) Establish domino property: 8k ≥ 0 Sk) Sk+1 • By induction on n, we conclude that: 8k ≥ 0 Sk

THEOREM: The sum of the first n odd numbers is n2.

Theorem? The sum of the first n numbers is ½n(n+1).

Theorem?The sum of the first n numbers is ½n(n+1). Try it out on small numbers! 1 = 1 =½1(1+1). 1+2 = 3 =½2(2+1). 1+2+3 = 6 =½3(3+1). 1+2+3+4 = 10 =½4(4+1).

Theorem?The sum of the first n numbers is ½n(n+1). = 0 =½0(0+1). 1 = 1 =½1(1+1). 1+2 = 3 =½2(2+1). 1+2+3 = 6 =½3(3+1). 1+2+3+4 = 10 =½4(4+1).

Notation:0 = 0 n= 1 + 2 + 3 + . . . + n-1 + n Let Snbe the statement “n =n(n+1)/2”

Sn´ “n =n(n+1)/2”Use induction to prove k ≥ 0, Sk Establish “Base Case”: S0. 0=The sum of the first 0 numbers = 0. Setting n=0, the formula gives 0(0+1)/2 = 0. Establish that k ≥ 0, Sk) Sk+1 “Inductive Hypothesis” Sk: k=k(k+1)/2 k+1 = k + (k+1) = k(k+1)/2 + (k+1) [Using I.H.] = (k+1)(k+2)/2 [which proves Sk+1]

Theorem: The sum of the first n numbers is ½n(n+1).

Primes: A natural number n>1 is called prime if it has no divisors besides 1 and itself. n.b. 1 is not considered prime.

Theorem:? Every natural number>1 can be factored into primes. Sn “n can be factored into primes” Base case: 2 is prime  S2 is true.

Trying to prove Sk-1) Sk • How do we use the fact Sk-1´ “k-1 can be factored into primes” • to prove thatSk´ “k can be factored into primes” • Hmm!?

This illustrates a technical point about using and defining mathematical induction.

Theorem:? Every natural number>1 can be factored into primes. A different approach: Assume 2,3,…,k-1 all can be factored into primes. Then show that k can be factored into primes.

All Previous InductionTo Prove k, Sk Establish Base Case: S0 Establish that k, Sk) Sk+1 Let k be any natural number. Induction Hypothesis: Assume j<k, Sj Use that to derive Sk

Also called “Strong Induction”

“All Previous” InductionCan Be Repackaged AsStandard Induction Define Ti = j ≤ i, Sj Establish “Base Case”: T0 Establish that k, Tk) Tk+1 Let k be any number. Assume Tk-1 Prove Tk Establish “Base Case”: S0 Establish that k, Sk) Sk+1 Let k be any number. Assume j<k, Sj Prove Sk

And there are more ways to do inductive proofs

Aristotle’s Contrapositive • Let S be a sentence of the form “A ) B”. • The Contrapositive of S is the sentence “B )A”. • A ) B: When A is true, B is true. • B )A: When B is false, A is false.

Aristotle’s Contrapositive • Logically equivalent: • A B “A)B” “B )A”. • False False True True • False True True True • True False False False • True True True True

Contrapositive or Least Counter-Example Induction to Prove k, Sk Establish “Base Case”: S0 Establish that k, Sk) Sk+1 Let k>0 be the least number such that Sk is false. Prove that Sk)Sk-1 Contradiction of k being the least counter-example!

Least Counter-Example Induction to Prove k, Sk Establish “Base Case”: S0 Establish that k, Sk) Sk+1 Assume that Sk is the least counter-example. Derive the existence of a smaller counter-example Sj (for j < k)

Rene Descartes [1596-1650]“Method Of Infinite Decent” Show that for any counter-example you find a smaller one. If a counter-example exists there would be an infinite sequence of smaller and smaller counter examples.

Each number > 1 has a prime factorization. • Let n be the least counter-example. • Hence n is not prime  so n = ab. If both a and b had prime factorizations, then n would too. • Thus a or b is a smaller counter-example.

Inductive reasoning is the high level idea: “Standard” Induction “All Previous” Induction “Least Counter-example” all just different packaging.

Euclid’s theorem on the unique factorization of a number into primes. Assume there is a least counter-example. Derive the existence of a smallercounter-example.

Theorem: Each natural has a unique factorization into primes written in non-decreasing order. • Let n be the least counter-example. n has at least two ways of being written as a product of primes: • n = p1 p2 .. pk = q1 q2 … qt • The p’s must be totally different primes than the q’s or else we could divide both sides by one of a common prime and get a smaller counter-example. Without loss of generality, assume p1 > q1 .

Induction: One Step At A Time